Let $g(x)=\sqrt{2x-4}$ and let $c$ be the number that satisfies the Mean Value Theorem for $g$ on the interval $2\leq x\leq10$. What is $c$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $2.25$ (Choice B) B $3.75$ (Choice C) C $4$ (Choice D) D $6$
According to the Mean Value Theorem, there exists a number $c$ in the open interval $2<x<10$ such that $g'(c)$ is equal to the average rate of change of $g$ over the interval: $g'(c)=\dfrac{g(10)-g(2)}{(10)-(2)}$ First, let's find that average rate of change: $\dfrac{g(10)-g(2)}{(10)-(2)}=\dfrac{4-0}{8}={\dfrac{1}{2}}$ Now, let's differentiate $g$ and find the $x$ -value for which $g'(x)={\dfrac{1}{2}}$. $g'(x)=\dfrac{1}{\sqrt{2x-4}}$ The solution of $g'(x)=\dfrac{1}{2}$ is $x=4$. $x=4$ is indeed within the interval $2<x<10$. In conclusion, $c=4$.